Optimal. Leaf size=316 \[ \frac {\left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {\left (5 a^3 B+30 a^2 A b-40 a b^2 B-16 A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 \sqrt {b} d}-\frac {(-b+i a)^{5/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {(b+i a)^{5/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d} \]
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Rubi [A] time = 3.07, antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3607, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ \frac {\left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {\left (30 a^2 A b+5 a^3 B-40 a b^2 B-16 A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 \sqrt {b} d}-\frac {(-b+i a)^{5/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {(b+i a)^{5/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d} \]
Antiderivative was successfully verified.
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Rule 63
Rule 93
Rule 205
Rule 206
Rule 208
Rule 217
Rule 3607
Rule 3647
Rule 3655
Rule 6725
Rubi steps
\begin {align*} \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{3} \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\frac {3}{2} a (2 a A-b B)+3 \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {3}{2} b (2 A b+3 a B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {\int \frac {\sqrt {a+b \tan (c+d x)} \left (-\frac {3}{4} a b (2 A b+3 a B)+6 b \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac {3}{4} b \left (14 a A b+5 a^2 B-8 b^2 B\right ) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{6 b}\\ &=\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {\int \frac {-\frac {3}{8} a b \left (18 a A b+11 a^2 B-8 b^2 B\right )+6 b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)+\frac {3}{8} b \left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{6 b}\\ &=\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {3}{8} a b \left (18 a A b+11 a^2 B-8 b^2 B\right )+6 b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x+\frac {3}{8} b \left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 b d}\\ &=\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {\operatorname {Subst}\left (\int \left (\frac {3 b \left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right )}{8 \sqrt {x} \sqrt {a+b x}}-\frac {6 \left (b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{6 b d}\\ &=\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {\operatorname {Subst}\left (\int \frac {b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {\operatorname {Subst}\left (\int \left (\frac {i b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {i b \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-b \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 d}\\ &=\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {\left ((a-i b)^3 (A-i B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {\left ((a+i b)^3 (A+i B)\right ) \operatorname {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 d}\\ &=\frac {\left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 \sqrt {b} d}+\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {\left ((a-i b)^3 (A-i B)\right ) \operatorname {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\left ((a+i b)^3 (A+i B)\right ) \operatorname {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=-\frac {(i a-b)^{5/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (30 a^2 A b-16 A b^3+5 a^3 B-40 a b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{8 \sqrt {b} d}+\frac {(i a+b)^{5/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (14 a A b+5 a^2 B-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{8 d}+\frac {(2 A b+3 a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\\ \end {align*}
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Mathematica [A] time = 4.56, size = 345, normalized size = 1.09 \[ \frac {3 \left (5 a^2 B+14 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {3 \sqrt {a} \left (5 a^3 B+30 a^2 A b-40 a b^2 B-16 A b^3\right ) \sqrt {\frac {b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}+24 \sqrt [4]{-1} (-a+i b)^{5/2} (A-i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+24 \sqrt [4]{-1} (a+i b)^{5/2} (A+i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+6 (3 a B+2 A b) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+8 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{24 d} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.37, size = 2655805, normalized size = 8404.45 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\tan \left (d x + c\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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