Optimal. Leaf size=268 \[ \frac {(a+i b)^2 (i A-B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {\left (12 a A b+3 a^2 B-8 b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 \sqrt {b} d}+\frac {(i a+b)^{3/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d} \]
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Rubi [A]
time = 1.68, antiderivative size = 268, 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 = {3688, 3728,
3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} \frac {\left (3 a^2 B+12 a A b-8 b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 \sqrt {b} d}+\frac {(a+i b)^2 (-B+i A) \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {(5 a B+4 A b) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {(b+i a)^{3/2} (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 {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 95
Rule 211
Rule 212
Rule 214
Rule 223
Rule 3688
Rule 3728
Rule 3736
Rule 6857
Rubi steps
\begin {align*} \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{2} \int \frac {\sqrt {\tan (c+d x)} \left (\frac {1}{2} a (4 a A-3 b B)+2 \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {1}{2} b (4 A b+5 a B) \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\int \frac {-\frac {1}{4} a b (4 A b+5 a B)+2 b \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac {1}{4} b \left (12 a A b+3 a^2 B-8 b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 b}\\ &=\frac {(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{4} a b (4 A b+5 a B)+2 b \left (a^2 A-A b^2-2 a b B\right ) x+\frac {1}{4} b \left (12 a A b+3 a^2 B-8 b^2 B\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac {(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\text {Subst}\left (\int \left (\frac {b \left (12 a A b+3 a^2 B-8 b^2 B\right )}{4 \sqrt {x} \sqrt {a+b x}}-\frac {2 \left (b \left (2 a A b+a^2 B-b^2 B\right )-b \left (a^2 A-A b^2-2 a b B\right ) x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac {(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {\text {Subst}\left (\int \frac {b \left (2 a A b+a^2 B-b^2 B\right )-b \left (a^2 A-A b^2-2 a b 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 (12 a A b+3 a^2 B-8 b^2 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac {(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {\text {Subst}\left (\int \left (\frac {b \left (a^2 A-A b^2-2 a b B\right )+i b \left (2 a A b+a^2 B-b^2 B\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {-b \left (a^2 A-A b^2-2 a b B\right )+i b \left (2 a A b+a^2 B-b^2 B\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (12 a A b+3 a^2 B-8 b^2 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 d}\\ &=\frac {(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\left ((a-i b)^2 (A-i B)\right ) \text {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)^2 (A+i B)\right ) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (12 a A b+3 a^2 B-8 b^2 B\right ) \text {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 )}{4 d}\\ &=\frac {\left (12 a A b+3 a^2 B-8 b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 \sqrt {b} d}+\frac {(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\left ((a-i b)^2 (A-i B)\right ) \text {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)^2 (A+i B)\right ) \text {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 {(a+i b)^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 )}{\sqrt {i a-b} d}+\frac {\left (12 a A b+3 a^2 B-8 b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 \sqrt {b} d}+\frac {(i a+b)^{3/2} (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 {(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\\ \end {align*}
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Mathematica [A]
time = 1.63, size = 290, normalized size = 1.08 \begin {gather*} \frac {-4 \sqrt [4]{-1} (-a+i b)^{3/2} (A-i B) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 \sqrt [4]{-1} (a+i b)^{3/2} (A+i B) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+(4 A b+5 a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 b B \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} \left (12 a A b+3 a^2 B-8 b^2 B\right ) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{4 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.51, size = 2400957, normalized size = 8958.79 \[\text {output too large to display}\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\tan {\left (c + d x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \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 )}^{3/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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