Optimal. Leaf size=240 \[ -\frac {(-b+i a)^{5/2} (-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (a B+2 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {(b+i a)^{5/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
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Rubi [A] time = 2.02, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3605, 3645, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ -\frac {(-b+i a)^{5/2} (-B+i A) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (a B+2 A b) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {(b+i a)^{5/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 63
Rule 93
Rule 205
Rule 206
Rule 208
Rule 217
Rule 3605
Rule 3645
Rule 3655
Rule 6725
Rubi steps
\begin {align*} \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\sqrt {a+b \tan (c+d x)} \left (\frac {3}{2} a (2 A b+a B)-\frac {3}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac {3}{2} b^2 B \tan ^2(c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4}{3} \int \frac {-\frac {3}{4} a \left (a^2 A-3 A b^2-3 a b B\right )-\frac {3}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+\frac {3}{4} b^3 B \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \operatorname {Subst}\left (\int \frac {-\frac {3}{4} a \left (a^2 A-3 A b^2-3 a b B\right )-\frac {3}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x+\frac {3}{4} b^3 B x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{3 d}\\ &=-\frac {2 a (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \operatorname {Subst}\left (\int \left (\frac {3 b^3 B}{4 \sqrt {x} \sqrt {a+b x}}-\frac {3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x\right )}{4 \sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{3 d}\\ &=-\frac {2 a (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\operatorname {Subst}\left (\int \frac {a^3 A-3 a A b^2-3 a^2 b B+b^3 B+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (b^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {2 a (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\operatorname {Subst}\left (\int \left (\frac {-3 a^2 A b+A b^3-a^3 B+3 a b^2 B+i \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 {3 a^2 A b-A b^3+a^3 B-3 a b^2 B+i \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 )}{d}+\frac {\left (2 b^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {2 a (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {\left (2 b^3 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 )}{d}-\frac {\left ((a-i b)^3 (i A+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 (-3 a^2 A b+A b^3-a^3 B+3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\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 {2 b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\left ((a-i b)^3 (i A+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 (-3 a^2 A b+A b^3-a^3 B+3 a b^2 B+i \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\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} (i A-B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {2 b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\\ \end {align*}
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Mathematica [C] time = 39.78, size = 139636, normalized size = 581.82 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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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 = 1.12, size = 2651450, normalized size = 11047.71 \[ \text {output too large to display} \]
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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
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 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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